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AACIMP 2009 Summer School lecture by Boris Goldengorin. "Logistics" course. 3rd - 4th hour. Part 2.

AACIMP 2009 Summer School lecture by Boris Goldengorin. "Logistics" course. 3rd - 4th hour. Part 2.

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Logistics. Example Logistics. Example Presentation Transcript

  • Operations Research 1 Chapter 3 Introduction to Operations Research Hillier & Liebermann Eighth Edition McGrawHill Operations Research 1 – 06/07 – Chapter 3 1
  • Contents • Linear Programming – Example – Formulations – Graphical solution – The linear programming model – Additional examples Operations Research 1 – 06/07 – Chapter 3 2
  • The Wyndor Glass Co. Example • Three plants for glass products (windows and doors): – Aluminium frames and hardware are made in Plant 1. – Wood frames are made in Plant 2. – Plant 3 produces the glass and assembles the products. • Capacity is released voor new products: – Two products are an option: product 1 and product 2. • Product 1 requires capacity in Plants 1 and 3 only. • Product 2 requires capacity in Plants 2 and 3 only. • Both products need capacity in plant 3. What is the most profitable mix to produce? Operations Research 1 – 06/07 – Chapter 3 3
  • Formulation as a Linear Programming Problem X1 = number of batches of product 1 produced per week X2 = number of batches of product 2 produced per week Z = total profit per week (in thousands of dollars) from producing these two products Table 3.1 Production Time per batch, Hours Product Production Time 1 2 Available per Week, Hours Plant 1 1 0 4 2 0 2 12 X1 and X2 are the decision variables for the model. Using the bottom row of table 3.1 we obtain Z = 3X1 + 5X3 3 2 2 18 Operations Research batch – Chapter 3 Profit per 1 – 06/07 $ 3,000 $ 5,000 4
  • Mathematical formulation of the LP problem Maximize Z = 3X1 + 5X2 Subject to restrictions X1 ≤4 2X2 ≤ 12 3X1 + 2X2 ≤ 18 And X1 ≥ 0, X2 ≥ 0. Operations Research 1 – 06/07 – Chapter 3 5
  • Graphical Solution figure 3.1 Figure 3.1 Shaded area shows the X1 ≤ 4 values of (X1,X2) allowed by: X2 ≥ 0 X1 ≥ 0, X2 ≥ 0, X1 ≤ 4 X1 ≥ 0 Operations Research 1 – 06/07 – Chapter 3 6
  • Graphical Solution figure 3.2 Figure 3.2 3X1 + 2X2 = 18 Shaded area shows the X1 =4 set of permissible values of (X1,X2), called the 2X2 = 12 feasible region. Operations Research 1 – 06/07 – Chapter 3 7
  • Graphical Solution figure 3.3 Z = 36 = 3X1 + 5X2 (2,6) Z = 20 = 3X1 + 5 X2 Z = 10 = 3X1 + 5X2 Operations Research 1 – 06/07 – Chapter 3 8
  • Terminology for Linear Programming TABLE 3.2 Prototype Example General Problem Production capacities of plants Resources 3 plants m resources Production of products Activities 2 products n activities Production rate of product j, xj Level of activity j, Xj Profit Z Objective function (Overall measure of performance Z) Operations Research 1 – 06/07 – Chapter 3 9
  • Data needed for a LP Model TABLE 3.3 Data needed for a linear programming model involving the allocation of resources to activities Resource Usage per Unit of Activity Activity Amount of Resource Resource 1 2 … n available 1 a11 a12 … a1n b1 2 a21 a22 … a2n b2 . . . … … … … . m am1 am2 … amn bm Contribution to 1 – 06/071 – Chapter23 Operations Research c c … cn 10 Z per unit of
  • Symbols used as notation of the various components commonly used for a LP Model • Below certain Symbols are listed, along with their interpretation of the general problem. – Z = Value of overall measure of performance. – xj = level of activity j (for j = 1, 2, …, n). – cj = increase in Z that would result from each unit increase in level of activity j. – bi = amount of resource i that is available for allocation to activities (for i = 1, 2, …, m). – aij = amount of resource i consumed by each other unit of acivity j. • The model poses the problem in terms of making decisions about the levels of the activitys, so x1, x2, …, xn are called the decision variables. In Table 3.3 the values of ci, bi, and aij (for i = 1, 2, …, m and j = 1, 2, … , n) are the input constants also referred to as the parameters of the model. Operations Research 1 – 06/07 – Chapter 3 11
  • A standard form of the model • We can now formulate the mathematical model for this general problem of allocation resources to activities. The model is to select the values of x1, x2, …, xn so as to Maximize Z = c1x1 + c2x2 + … + cnxn, subject to restrictions a11x1 + a12x2 + … + a1nxn ≤ b1 a21x1 + a22x2 + … + a2nxn ≤ b2 ………………………… am1x1 + am2x2 + … + amnxn ≤ bm, and x1 ≥ 0, x2 ≥ 0 , …, xn ≥ 0 Operations Research 1 – 06/07 – Chapter 3 12
  • Summary of terminology in LP 1 objective function = the function being maximized: c1x1 + c2x2 + … + cnxn. constraints = restrictions for the objective function. functional constraints = the first m constraints with a function of all the variables (ai1x1 + ...+ ainxn) on the left- hand side. nonnegativity constraints = the xj ≥ 0 restrictions Feasible solution = a solution for which all the constraints are satisfied. Infeasible solution = a solution for which at least one constraint is violated. Feasible region = the collection of all feasible solutions. Operations Research 1 – 06/07 – Chapter 3 13
  • Summary of terminology in LP 2 Optimal solution = a feasible solution that has the most favorable value of the objective function. Most favorable value = the largest value if the objective function is to be maximized or the smallest value if the objective function is to be minimized. It is possible for a problem to have no feasible solutions, multiple optimal solutions. If there are no optimal solutions we speak of an unbounded Z or an unbounded objective Operations Research 1 – 06/07 – Chapter 3 14
  • Figure 3.4 Maximize Z = 3x1 + 5x2, Subject to X1 ≤ 4 3x1 + 5x2 ≥ 50 2x2 ≤ 12 3x1 + 2x2 ≤ 18 3x1 + 5x2 ≥ 50 And x1 ≥ 0, x2 ≥ 0 2x2 ≤ 12 3x1 + 2x2 ≤ 18 x1 ≤ 0 Figure 3.4 X1 ≤ 4 The Wyndor Glass Co. x2 ≥ 0 Problem would have no solutions if the constraint 3x1 + 5x2 ≥ 50 were added to the problem. Operations Research 1 – 06/07 – Chapter 3 15
  • Figure 3.5 Maximize Z = 3x1 + 2x2, Subject to X1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 And x1 ≥ 0, x2 ≥ 0 Every point on this darker line segment is optimal, each with Z = 18 Feasible Figure 3.5 region The Wyndor Glass Co. Problem would have multiple solutions if the objective function were changed to Z = 3x1 + 2x2 Operations Research 1 – 06/07 – Chapter 3 16
  • Figure 3.6 (4,10), Z = 62 (4,8), Z = 52 Maximize Z = 3x1 + 5x2 Figure 3.6 Subject to x1 ≤ 4 (4,6), Z = 42 The Wyndor Glass Co. And x1 ≥ 0, x2 ≥ 0 Problem would have no Feasible optimal solutions if the region (4,4), Z = 32 only functional constraint were x1 ≤ 4, because x2 then could (4,2), Z = 22 be increased indefinitely in the feasible region without ever reaching the maximum value of Z = 3x1 + 5x2 Operations Research 1 – 06/07 – Chapter 3 17
  • Figure 3.7 (0,6) (2,6) (4,3) Figure 3.7 The five dots are the five (4,0) CPF solutions for the Wyndor Glass Co. (0,0) Problem. Operations Research 1 – 06/07 – Chapter 3 18
  • Assumptions of LP Proportionality assumption: The contribution of each activity to the value of the objective function Z is proportional to the level of the activity xj, as represented by the cjxj term in the objective function. Similarly, the contribution of each activity to the left-hand side of each functional constraint is proportional to the level of activity xj, as represented by the aijxj term in the constraint. Consequently, this assumption rules out any exponent other than 1 for any variable in any term of any function (whether the objective function or the function on the left- hand side of a functional constraint) in a linear programming model. Operations Research 1 – 06/07 – Chapter 3 19
  • Graphs of Table 3.4 x2 20 18 18 16 14 Proportionality Satisfied 12 12 11 Case 1 10 8 8 Case 2 7 6 6 6 5 Case 3 4 3 2 2 0 0 0 1 2 3 4 x1 In case 1 through 3 proportionality is violated. Operations Research 1 – 06/07 – Chapter 3 20
  • Additivity assumption & Table 3.5 Additivity Assumption: Every function in a linear programming model (whether the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities. • TABLE 3.5 Examples of satisfying or violating additivity for the objective function Value of Z Additivity Violated (x1,x2) Additivity Satisfied Case 1 Case 2 (1,0) 3 3 3 (0,1) 5 5 5 (1,1) 8 9 7 Operations Research 1 – 06/07 – Chapter 3 21
  • Table 3.6 • TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint Amount of Resource Used Additivity Violated (x1,x2) Additivity Satisfied Case 3 Case 4 (2,0) 6 6 6 (0,3) 6 6 6 (2,3) 12 15 10.8 Operations Research 1 – 06/07 – Chapter 3 22
  • Divisibility assumption Divisibility Assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values. Since each decision variable presents the level of some activity, it is being assumed that the activities can be run at fractional levels. Operations Research 1 – 06/07 – Chapter 3 23
  • Certainty assumption Certainty assumption: The value assigned to each parameter of a linear programming model is assumed to be a known constant. Operations Research 1 – 06/07 – Chapter 3 24
  • Regional Planning: The southern confederation of kibbutzim • TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim Kibbutz Usable Land (Acres) Water Allocation (Acre Feet) 1 400 600 2 600 800 3 300 375 • TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim Crop Maximum Water Consumption Net Return ($ / Acre) Quota (Acres) (Acre Feet/ Acre) Sugar beets 600 3 1,000 Cotton 500 2 750 Sorghum 325 1 250 Operations Research 1 – 06/07 – Chapter 3 25
  • Regional Planning (continued): Table 3.10 • TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem Allocation (Acres) Kibbutz Crop 1 2 3 Sugar beets X1 X2 X3 Cotton X4 X5 X6 Sorghum X7 X8 X9 Operations Research 1 – 06/07 – Chapter 3 26